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Re: Is (x – 2)^2 > x^2? (1) x^2 > x (2) 1/x > 0 [#permalink]
Need some help here.

Is (x - 2)^2 > x^2?

Let's work on the stem first:

Is (x-2)^2>x^2? --> Is x^2-4x+4>x^2? --> Is x<1? So this is what the question is basically asking.

(1) x^2>x --> x(x-1)>0 --> x<0 OR x>1. Two ranges, not sufficient. As per me it should be x>0 and x>1. I know this is wrong as I tried with 1/2 as a value of x which falsifies the equation.

Per my understanding,
if the equation were x(x-1) = 0, x can have two values i.e. 0 & 1. Please help me understand if I am missing out something.

(2) \frac{1}{x}>0 --> x>0. Not sufficient.

(1)+(2) Intersection of the ranges from (1) and (2) is x>1. Hence the answer to the question is no. Sufficient.

Answer: C.
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Re: Is (x – 2)^2 > x^2? (1) x^2 > x (2) 1/x > 0 [#permalink]
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honey86 wrote:
Need some help here.

Is (x - 2)^2 > x^2?

Let's work on the stem first:

Is (x-2)^2>x^2? --> Is x^2-4x+4>x^2? --> Is x<1? So this is what the question is basically asking.

(1) x^2>x --> x(x-1)>0 --> x<0 OR x>1. Two ranges, not sufficient. As per me it should be x>0 and x>1. I know this is wrong as I tried with 1/2 as a value of x which falsifies the equation.

Per my understanding,
if the equation were x(x-1) = 0, x can have two values i.e. 0 & 1. Please help me understand if I am missing out something.

(2) \frac{1}{x}>0 --> x>0. Not sufficient.

(1)+(2) Intersection of the ranges from (1) and (2) is x>1. Hence the answer to the question is no. Sufficient.

Answer: C.


x>0 and x>1 does not make any sense.

x(x-1)>0 --> the "roots" are 0 and 1, this gives us 3 ranges:

x<0;
0<x<1;
x>1.

Next, test some extreme value for x: if x is some large enough number, say 10, then both multiples will be positive which gives the positive result for the whole expression, so when x>1 the expression is positive. Now the trick: as in the 3rd range expression is positive then in the 2nd it'll be negative and finally in the 1st it'll be positive again: + - + . So, the ranges when the expression is positive are: x<0 and x>1.

Check this for more: x2-4x-94661.html#p731476

Hope it helps.
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Re: Is (x – 2)^2 > x^2? (1) x^2 > x (2) 1/x > 0 [#permalink]
Got it!
Thanks a lot Bunuel. You have explained it very well on the other link that you posted.

Will work on more such problems.
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Re: Is (x – 2)^2 > x^2? (1) x^2 > x (2) 1/x > 0 [#permalink]
How about this case?
Since its not mentioned x an is integer or a fraction, let x = 3/2
So stmt becomes (3/2 - 2)^2 > (3/2) which is false.

Ans C is when 0 < x < 1
But when x > 1, then ans should be E
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Re: Is (x – 2)^2 > x^2? (1) x^2 > x (2) 1/x > 0 [#permalink]
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Sailesh1986 wrote:
How about this case?
Since its not mentioned x an is integer or a fraction, let x = 3/2
So stmt becomes (3/2 - 2)^2 > (3/2) which is false.

Ans C is when 0 < x < 1
But when x > 1, then ans should be E


x cannot be less than or equal to 1, because when we combine the statements we get that x>1.
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Re: Is (x – 2)^2 > x^2? (1) x^2 > x (2) 1/x > 0 [#permalink]
Why not A?

In 1, if x^2>x, then wouldn't x>1?

Originally posted by US09 on 27 Jan 2018, 06:46.
Last edited by US09 on 27 Jan 2018, 07:57, edited 1 time in total.
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Re: Is (x – 2)^2 > x^2? (1) x^2 > x (2) 1/x > 0 [#permalink]
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tamal99 wrote:
Is (x – 2)^2 > x^2?

(1) x^2 > x
(2) 1/x > 0


\((x – 2)^2 > x^2.....x^2-4x+4>x^2....4x<4....x<1\)
So Q basically asks us - Is x<1?

(1) \(x^2 > x\)..
\(x^2 > x....x^2-x>0...x(x-1)>0\)..
so if x>0, x-1>0 or x>1...NO
if x<0, x-1<0 or x<1...YES
insuff

(2) \(\frac{1}{x} > 0\)..
\(\frac{1}{x} > 0\)..
this tells us that x >0
insuff

combined
x>0, so x>1
ans is NO
suff
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Re: Is (x – 2)^2 > x^2? (1) x^2 > x (2) 1/x > 0 [#permalink]
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urvashis09 wrote:
Why not A?

In 1, if x^2>x, then wouldn't x>1?


Hi Urvashi

If x^2 > x , then it could mean two things:
Either x > 1 Or
x < 0 (x can take any negative value which would make its square positive, positive is always greater than negative).
Thats why first statement alone is NOT sufficient.
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Re: Is (x – 2)^2 > x^2? (1) x^2 > x (2) 1/x > 0 [#permalink]
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kirankp wrote:
Is \((x – 2)^2 > x^2\)?


(1) \(x^2 > x\)

(2) \(\frac{1}{x} > 0\)


Target question: Is (x - 2)² > x²?
This is a great candidate for rephrasing the target question.

Take (x - 2)² > x² and expand the left side to get: x² - 4x + 4 > x²
Subtract x² from both sides to get: -4x + 4 > 0
Add 4x to both sides to get: 4 > 4x
Divide both sides by 4 to get: 1 > x
REPHRASED target question: Is x < 1?

Aside: the video below has tips on rephrasing the target question

Statement 1: x² > x
Subtract x from both sides of the inequality to get: x² - x > 0
Factor to get: x(x - 1) > 0
This means that EITHER x < 0 OR x > 1
So there are two possible cases to consider.
case a: If x < 0, then it IS the case that x < 1
case b: If x > 1, then it is NOT the case that x < 1
Since we cannot answer the REPHRASED target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: (1/x) > 0
If 1 divided by x equals some POSITIVE value, we can conclude that x is POSITIVE
If x is POSITIVE, then x could be greater than 1, or x could be less than 1
Since we cannot answer the REPHRASED target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 1 tells us that that EITHER x < 0 OR x > 1
Statement 2 tells us that x is POSITIVE
So, we can eliminate the possibility that x < 0
This means it MUST be the case that x > 1
So, we can conclude that x is NOT less than 1
Since we can answer the REPHRASED target question with certainty, the combined statements are SUFFICIENT

Answer: C

Cheers,
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Re: Is (x – 2)2 > x2? (1) x2 > x (2) 1/x > 0 [#permalink]
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Is \( (x – 2)^2 > x^2\)?
\((x^2-4x+4)>(x)^2………..4x<4………x<1\)

So we are looking for whether x is less than 1.

(1) \(x^2 > x…..x(x-1)>0\)
If x>0, then x-1>0 or x>1….x>1
If x<0, then x-1<0 or x<1….x<0
We cannot say whether x<1.
Insufficient

(2) \(\frac{1}{x} > 0\)
Since 1>0, the denominator x is also >0.
Again, we cannot say whether x<1.
Insufficient


Combined
x>0, and from statement I we get x>1.
Sufficient as answer is NO.


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Re: Is (x 2)^2 > x^2? (1) x^2 > x (2) 1/x > 0 [#permalink]
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